Gujarat Board GSEB Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1.

Which of the following options is true and why?

y = 3x + 5

(i) a unique solution

(ii) only two solutions

(iii) infinitely many solutions

Solution:

Option (iii) is true y = 3x + 5 has infinitely many solutions.

Reason: for every real value of x, there will be a corresponding value of y and vice versa.

Question 2.

Write four solutions for each of the following equations.

1. 2x + y = 7

2. πx + y = 9

3. x = 4y

Solution:

1. 2x + y = 7

⇒ y = 7 – 2x

Taking values of x = 0, 1, 2, 3

Putting x = 0, y = 7 – 2 x 0 = 7

Putting x = 1, y = 7 – 2 x 1 = 7 – 2 = 5

Putting x = 2, y = 7 – 2x

Putting x = 3, y = 7 – 2 x 3 = 7 – 6 = 1

Therefore four solutions are

(0, 7), (1, 5), (2, 5) and (3, 1)

2. πx + y = 9

⇒ y = 9 – πx

Now putting the values of x = 0, 1, 2, 3

Putting x = 0, y = 9 – π x 0

y = 9

Putting x = 1, y = 9 – π

Putting x = 2, y = 9 – πx = 9 – π x 2

y = 9 – 2π

Putting x = 3, y = 9 – πx

y = 9 – π x 3

⇒ y = 9 – 3π

Therefore four solutions are

(0, 9), (1, 9 – π ), (2, 9 – 2π) and (3, 9 – 3π)

3. x = 4y

Putting value ofy in given eqn, we get value of x.

Let’s take values of y = 0, 1, 2, 3

Puttmg y = 0, x = 0

Putting y = 1, x = 4 x 1

x = 4

Putting y = 2, x = 4 x 2

x = 8

Putting y = 3, x = 4 x 3

x = 12

Therefore four solutions are

(0, 0), (4, 1), (8, 2) and (12, 3)

Question 3.

Check which of the following are solutions of the equation x – 2y = 4 and which are not?

1. (0, 2)

2. (2, 0)

3. (4, 0)

4. \((\sqrt{2}, 4 \sqrt{2})\)

5. (1,1)

Solution:

1. x – 2y = 4

(0, 2) will be the solution of given equation

if they will satisfy the given equation.

x – 2y = 4

Puttingx = 0 and y = 2 in LHS,

0 – 2 x 2 = -4

⇒ LHS = RHS

∴ (0,2) is not the solution of given equation.

2. x – 2y = 4

(2, 0) will be the solution of given equation

if they satisfy the equation.

Putting x = 2 and y = 0 in LHS,

⇒ 2 – 2 x 0 = 2

LHS = RHS

∴ (2,0) is not the solution of given equation.

3. x – 2y = 4

(4, 0) will be the solution of given equation

if they satisfy the equation.

Putting x = 4 and y = 0 in LHS,

4- 2 x 0 = 4

LHS = RHS

∴ (4, 0) is the solution of given equation.

4. x – 2y = 4

\((\sqrt{2}, 4 \sqrt{2})\) will be the solution of given equation if they satisfy the equation.

Putting x = \(\sqrt{2}\) and y = 4\(\sqrt{2}\) in LHS

\(\sqrt{2}\) – 2 x 4\(\sqrt{2}\)

\(\sqrt{2}\) – 8\(\sqrt{2}\) = -7\(\sqrt{2}\)

⇒ LHS = RHS

\((\sqrt{2}, 4 \sqrt{2})\) is not the solution of given equation.

5. x – 2y = 4

(1, 1) will be the solution of given equation

if they satisfy the equation.

Putting x = 1 and y = 1 in LHS,

1 – 2 x 1= -1

⇒ LHS = RHS

∴ (1, 1) is not the solution of given equation.

Question 4.

Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Solution:

Given 2x + 3y = k

x = 2 and y = 1 will be the solution of given equation if this point satisfies the equation.

2x + 3y = k

2 x 2 + 3 x 1 = k

4 + 3 = k

k = 7